Cutting along a simply connected hypersurface adds one free generator

Put $U=M\setminus S$ and $V=N\cup T$. These are path-connected open sets covering $M$. Their intersection consists of $N_{-}$, $N_{+}$, and the tube joining their end balls. Cover the intersection by the two side components with the respective half of the tube. Each set is path-connected and simply connected, and their overlap is a contractible middle tube. Hence [F1] gives ${\pi }_1(U\cap V)=1$.

Join the centers of the two end balls by an embedded arc $\alpha$ in the connected smooth $3$-manifold $N$, and let $Q$ be a thin open tubular neighborhood of $\alpha$ in $N$, enlarged at its ends to meet those balls. The open sets $N$ and $W=T\cup Q$ cover $V$. The set $W$ is a thickened loop formed by the two tubes $T$ and $Q$, so ${\pi }_1(W)\cong \mathbb{Z}$; $N\cap W$ is $Q$ together with the two end balls and is contractible. Since $N$ is simply connected, [F1] gives ${\pi }_1(V)\cong \mathbb{Z}$.

Apply [F1] to the open cover $M=U\cup V$. Step 1.1 gives ${\pi }_1(U\cap V)=1$, while ${\pi }_1(U)={\pi }_1(M\setminus S)$ and step 2.1 gives ${\pi }_1(V)=\mathbb{Z}$. The pushout is therefore the free product ${\pi }_1(M\setminus S)\ast \mathbb{Z}$. A generator of the second factor is represented by $\alpha$ followed by ${\nu }^{-1}$, proving the stated description.